0
$\begingroup$

I'm following Eisenbud-Harris Geometry of Schemes, and I am a little bit confused about how they define the notion of a reduced scheme. In the affine case, if $X = \text{Spec} \, R$, then setting $X_{\mathrm{red}} = \text{Spec}\, R_{\mathrm{red}}$ where $R_{\mathrm{red}}$ is the ring $R$ modulo its nil radical, we say that $X$ is reduced if $X = X_{\mathrm{red}}$.

For an arbitrary scheme $X$, they define a quasi-coherent sheaf called the nilradical, that assigns to each open subset $U$ of $X$, the nilradical of $\mathcal{O}_X(U)$, being $\mathcal{O}_X$ the structure sheaf of $X$, and then say that the scheme $X$ is reduced if its associated closed subscheme $X_{\mathrm{red}}$ is equal to $X$.

What does the statement "its associated closed subscheme $X_{\mathrm{red}}$" mean?

The definition Eisenbud and Harris give of a closed subscheme $Y$ of a scheme $X$ is a closed topological subspace $|Y| \subset |X|$ with a sheaf $\mathcal{O}_Y$ that is the quotient sheaf of the structure sheaf of $X$ by a quasi-coherent sheaf of ideals $\mathcal{J}$ such that the intersection of $Y$ with any affine open subset $U \subset X$ is the closed subschema associated to the ideal $\mathcal{J}(U)$.

Since we have a quasi-coherent sheaf $\mathcal{N}$, we can quotient the structure sheaf $\mathcal{O}_X$ of $X$ by $\mathcal{N}$. Is then
$$X_{\mathrm{red}} = (X, \mathcal{O}_X/\mathcal{N})\, ?$$ The problem is that the definition of a closed subscheme $Y$ of a scheme $X$ says that the topological space $|Y|$ has to be a closed subspace of the topological space $|X|$, but in this case it would be the same space and not a proper subspace.

$\endgroup$
  • 2
    $\begingroup$ Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however. $\endgroup$ – Sasha Nov 16 '18 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.